Final Exam Review Problems CHEM FINAL

Significant Figures

• Question 1 involves determining the correct number of significant figures in the result of a calculation: $(0.014371 m + 0.320 m) \ \times (1.04 m + 0.3 m)$

  • The first part of the equation, addition, requires us to consider decimal places. 0. 014371 has six decimal places while 0.320 has three decimal places. The answer should have three decimal places.

    • $0.014371 + 0.320 = 0.334371 \approx 0.334$

  • The second part of the equation, addition, requires us to consider decimal places. 1. 04 has two decimal places while 0.3 has one decimal place. The answer should have one decimal place.

    • $1.04 + 0.3 = 1.34 \approx 1.3$

  • Now, consider multiplication. $0.334 \times 1.3 = 0.4342 \approx 0.4$ This restricts us to one significant figure.

Unit Conversions

• Question 2 asks how many $nm^2$ is $1.35 cm^2$.

  • $1 cm = 1 \times 10^{-2} m$

  • $1 nm = 1 \times 10^{-9} m$

  • $1.35 cm^2 \times (\frac{1 \times 10^{-2} m}{1 cm})^2 \times (\frac{1 nm}{1 \times 10^{-9} m})^2 = 1.35 \times 10^{14} nm^2$

Atomic Structure and Isotopes

• Question 3 asks the number of protons, neutrons, and electrons in the most likely cation formed from barium-140.

  • Barium (Ba) is in group 2, so it forms a 2+ cation ($Ba^{2+}$).

  • Barium has an atomic number of 56, so it has 56 protons.

  • $Number \ of \ neutrons = mass \ number - atomic \ number$

    • $140 - 56 = 84 \ neutrons$

  • For a neutral atom, the number of electrons equals the number of protons. For a 2+ cation, there are two fewer electrons than protons.

    • $56 - 2 = 54 \ electrons$

• Question 4 presents nuclear compositions for atoms X, Y, Z, and R to identify isotopes.

  • Isotopes have the same number of protons but different numbers of neutrons.

  • X has 17 protons and 18 neutrons.

  • Y has 16 protons and 17 neutrons.

  • Z has 17 protons and 20 neutrons.

  • R has 16 protons and 18 neutrons.

  • X and Z have the same number of protons (17) but different numbers of neutrons, so they are isotopes.

Ionization Energy

• Question 5 deals with placing N, F, and As in order of increasing ionization energy.

  • Ionization energy increases across a period (from left to right) and decreases down a group.

  • N and F are in the same period, with F to the right of N, so F has higher ionization energy than N.

  • N and As are in the same group, with N above As, so N has higher ionization energy than As.

  • Therefore, the order of increasing ionization energy is As < N < F.

Quantum Numbers

• Question 6 asks about the maximum number of electrons that can have specific quantum numbers:

  • I. $n = 2, l = 1, m_l = -1$

    • When $n=2$ and $l=1$, this corresponds to the $2p$ orbitals.

    • $m_l = -1$ specifies one particular $2p$ orbital. Since each orbital can hold two electrons, the maximum number of electrons is 2.

  • II. $n = 3, m_l = 0$

    • For $n = 3$, $l$ can be 0, 1, or 2. For each value of $l$, there is a corresponding value of $m<em>l = 0$. For $l = 0$, there is one $m</em>l = 0$ ($3s$). For $l = 1$, there is one $m<em>l = 0$ ($3p$). For $l = 2$, there is one $m</em>l = 0$ ($3d$).

  • III. $n = 4, l = 2$

    • For $n = 4$ and $l = 2$, this corresponds to the $4d$ orbitals.

    • There are five $4d$ orbitals. Since each orbital can hold two electrons, $2 \times 5 = 10$ electrons

• Question 9 asks which set of four quantum numbers is allowed for an electron.

  • A. $n = 3, l = 2, m<em>l = -3, m</em>s = -\frac{1}{2}$

    • Not allowed because if $l = 2$, $m_l$ can only be -2, -1, 0, 1, or 2.

  • B. $n = 1, l = 1, m<em>l = 0, m</em>s = + \frac{1}{2}$

    • Not allowed because if $n = 1$, $l$ must be 0.

  • C. $n = 3, l = 1, m<em>l = -2, m</em>s = + \frac{1}{2}$

    • Not allowed because if $l = 1$, $m_l$ can only be -1, 0, or 1.

  • D. $n = 2, l = 1, m<em>l = 0, m</em>s = + 0$

    • Not allowed because ms can only be $+\frac{1}{2}$ or $-\frac{1}{2}$.

  • E. $n = 2, l = 1, m<em>l = -1, m</em>s = -\frac{1}{2}$

    • This set is allowed.

Isoelectronic Ions and Ionic Radius

• Question 7 involves isoelectronic ions Cl–, K+, and Ca2+ and asks to list them in order of decreasing ionic radius.

  • Isoelectronic species have the same number of electrons.

    • Cl– has 18 electrons (17 + 1).

    • K+ has 18 electrons (19 - 1).

    • Ca2+ has 18 electrons (20 - 2).

  • For isoelectronic ions, ionic radius decreases with increasing nuclear charge (more protons).

    • Cl– has 17 protons.

    • K+ has 19 protons.

    • Ca2+ has 20 protons.

  • Therefore, the order of decreasing ionic radius is Cl– > K+ > Ca2+.

Unpaired Electrons

• Question 8 asks how many unpaired electrons are present in the ground state of $Mn^{2+}$.

  • Manganese (Mn) has an atomic number of 25. Its electron configuration is $[Ar] 4s^2 3d^5$.

  • $Mn^{2+}$ loses two electrons, so the electron configuration is $[Ar] 3d^5$.

  • There are five 3d orbitals, and each of the five electrons occupies a separate orbital to minimize electron-electron repulsion (Hund's rule).

  • Therefore, there are 5 unpaired electrons.

Frequency of Light Emitted

• Question 10 asks to calculate the frequency of light emitted when the transition $n = 6 \rightarrow n = 3$ occurs in the hydrogen atom.

  • Use the Rydberg formula: $\frac{1}{\lambda} = R<em>H (\frac{1}{n</em>1^2} - \frac{1}{n<em>2^2})$ where $R</em>H$ is the Rydberg constant ($1.097 \times 10^7 m^{-1}$), $n<em>1$ is the final energy level, and $n</em>2$ is the initial energy level.

  • $\frac{1}{\lambda} = 1.097 \times 10^7 m^{-1} (\frac{1}{3^2} - \frac{1}{6^2}) = 1.097 \times 10^7 m^{-1} (\frac{1}{9} - \frac{1}{36}) = 1.097 \times 10^7 m^{-1} (\frac{4-1}{36}) = 1.097 \times 10^7 m^{-1} (\frac{3}{36}) = 1.097 \times 10^7 m^{-1} (\frac{1}{12})$

  • $\frac{1}{\lambda} = 9.1416 \times 10^5 m^{-1}$

  • $\lambda = \frac{1}{9.1416 \times 10^5 m^{-1}} = 1.094 \times 10^{-6} m$

  • $\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 m/s}{1.094 \times 10^{-6} m} = 2.74 \times 10^{14} s^{-1}$

Electron Configurations

• Question 11 asks you to select the atom with the incorrect electron configuration.

  • Cr: $[Ar]4s^13d^5$ (Correct - Chromium is an exception to Hund's rule)

  • Ni: $[Ar]4s^23d^8$ (Correct)

  • S: $[Ne]3s^23p^4$ (Correct)

  • Br: $[Ar]4s^23d^{10}4p^5$ (Correct)

  • Sn: $[Kr]5s^25d^{10}5p^2$ (Incorrect, should be $[Kr]5s^24d^{10}5p^2$)

Atomic Mass and Isotopes

• Question 12 A new compound was recently discovered and found to have an atomic mass of 342.38 amu. This element has two isotopes, the lighter of which has a mass of 340.91 amu and an abundance of 68.322%. What is the mass of the heavier isotope?

  • Let x be the mass of the heavier isotope, and let y be the abundance of the heavier isotope.

  • We know that y = 100% - 68.322% = 31.678%

  • $342.38 = (340.91 \times 0.68322) + (x \times 0.31678)$

  • $342.38 = 232.92 + 0.31678x$

  • $109.46 = 0.31678x$

  • $x = 345.55 \ amu$

Density and Number of Atoms

• Question 13 The density of silver is $10.5 g /cm^3$. How many Ag atoms are present in a silver bar that measures 0. 1000 m x 0. 0500 m x 0. 0100 m?

  • Volume of the silver bar = $0.1000 \times 0.0500 \times 0.0100 = 5.00 \times 10^{-5} m^3$

  • Convert $m^3$ to $cm^3$: $5.00 \times 10^{-5} m^3 \times (\frac{100 cm}{1 m})^3 = 5.00 \times 10^{-5} \times 10^6 cm^3 = 50.0 cm^3$

  • Mass of the silver bar = $Density \times Volume = 10.5 g/cm^3 \times 50.0 cm^3 = 525 g$

  • Convert grams to moles using the molar mass of silver (107.87 g/mol): $525 g \times (\frac{1 mol}{107.87 g}) = 4.867 mol$

  • Convert moles to atoms using Avogadro's number: $4.867 mol \times (\frac{6.022 \times 10^{23} atoms}{1 mol}) = 2.93 \times 10^{24} atoms$

Noble Gas Configuration

*Question 14 Which of the following has a noble gas configuration?
*Note that all of the options have a noble gas configuration.

• $P^{3-}$: Phosphorus has 15 electrons. $P^{3-}$ gains 3 electrons, resulting in 18 electrons which is the same as Argon.

• $Sr^{2+}$: Strontium has 38 electrons. Losing 2 electrons makes it isoelectronic with Krypton, which has 36 electrons.

• $Al^{3+}$: Aluminum has 13 electrons. $Al^{3+}$ loses 3 electrons, resulting in 10 electrons which is the same as Neon.

• $C^{4-}$: Carbon has 6 electrons. $C^{4-}$ gains 4 electrons, resulting in 10 electrons which is the same as Neon.

Atoms of Oxygen in a Compound

• Question 15, asks how many atoms of oxygen are contained in 47.6 g of $Al<em>2(CO</em>3)_3$ (Molar mass = 234 g/mol).

  • First, find the number of moles of the compound: $moles = \frac{mass}{molar \ mass} = \frac{47.6 g}{234 g/mol} = 0.2034 \ mol$

  • Determine the number of moles of oxygen in the compound. Each formula unit of $Al<em>2(CO</em>3)<em>3$ contains 9 oxygen atoms. $0.2034 \ mol \ Al</em>2(CO<em>3)</em>3 \times \frac{9 \ mol \ O}{1 \ mol \ Al<em>2(CO</em>3)_3} = 1.831 \ mol \ O$

  • Convert moles of oxygen to number of oxygen atoms using Avogadro's number: $1.831 \ mol \ O \times \frac{6.022 \times 10^{23} \ atoms}{1 \ mol} = 1.10 \times 10^{24} O \ atoms$

Ionic vs Covalent Compounds

• Question 16: When calcium reacts with nitrogen, it forms an ionic compound $Ca<em>3N</em>2$.

  • Calcium (Ca) is a metal and Nitrogen (N) is a nonmetal. Metals and nonmetals form ionic compounds.

    • Calcium forms a +2 ion ($Ca^{2+}$), and nitrogen forms a -3 ion ($N^{3-}$). Thus, three calcium ions are needed to balance two nitrogen ions.

Correct Chemical Formulas

• Question 17 asks to identify how many of the student's proposed formulas are correct.

  • Aluminum perchlorate: The correct formula is $Al(ClO<em>4)</em>3$. Aluminum has a +3 charge, and perchlorate has a -1 charge.

  • Cobalt(III) oxide: The correct formula is $Co<em>2O</em>3$. Cobalt(III) has a +3 charge, and oxygen has a -2 charge.

  • Sodium oxide: The correct formula is $Na_2O$. Sodium has a +1 charge, and oxygen has a -2 charge. The student's answer of NaO is incorrect.

  • Calcium phosphate: The correct formula is $Ca<em>3(PO</em>4)_2$. Calcium has a +2 charge, and phosphate has a -3 charge.

  • Phosphide ion: The correct formula is $P^{3-}$. The student's answer of $PO_3^{2-}$ is incorrect.

• Formulas that are correct are Aluminum perchlorate, calcium phosphate and Cobalt (III) oxide.

Empirical Formula

• Question 18 asks to determine the empirical formula of hydroxylamine nitrate, which contains 29.17 mass % N, 4.20 mass % H, and 66. 63 mass % O.

  • Assume 100 g of the compound, so we have 29.17 g N, 4.20 g H, and 66.63 g O.

  • Convert each mass to moles:

    • $29.17 g \ N \times \frac{1 \ mol \ N}{14.01 g \ N} = 2.082 \ mol \ N$

    • $4.20 g \ H \times \frac{1 \ mol \ H}{1.008 g \ H} = 4.167 \ mol \ H$

    • $66.63 g \ O \times \frac{1 \ mol \ O}{16.00 g \ O} = 4.164 \ mol \ O$

  • Divide each mole value by the smallest (2.082) to find the mole ratio:

    • N: $\frac{2.082}{2.082} = 1$

    • H: $\frac{4.167}{2.082} = 2$

    • O: $\frac{4.164}{2.082} = 2$

  • The empirical formula is $H<em>2NO</em>2$.

• Question 19 requires the determination of the empirical formula of a compound containing only C, H, and O.

  • Combustion of 5.000 g of the compound produces 11.88 g $CO<em>2$ and 2.835 g $H</em>2O$.

  • First, calculate the moles of C and H:

    • $moles \ of \ C = 11.88 g \ CO<em>2 \times \frac{1 \ mol \ CO</em>2}{44.01 g \ CO<em>2} \times \frac{1 \ mol \ C}{1 \ mol \ CO</em>2} = 0.270 \ mol \ C$

    • $moles \ of \ H = 2.835 g \ H<em>2O \times \frac{1 \ mol \ H</em>2O}{18.015 g \ H<em>2O} \times \frac{2 \ mol \ H}{1 \ mol \ H</em>2O} = 0.3147 \ mol \ H$

  • Calculate the mass of C and H in the original compound:

    • $mass \ of \ C = 0.270 \ mol \ C \times \frac{12.01 g \ C}{1 \ mol \ C} = 3.243 g \ C$

    • $mass \ of \ H = 0.3147 \ mol \ H \times \frac{1.008 g \ H}{1 \ mol \ H} = 0.3172 g \ H$

  • Calculate the mass of O in the original compound by subtracting the masses of C and H from the total mass of the compound:

    • $mass \ of \ O = 5.000 g - 3.243 g - 0.3172 g = 1.4398 g \ O$

  • Convert the mass of O to moles:

    • $moles \ of \ O = 1.4398 g \ O \times \frac{1 \ mol \ O}{16.00 g \ O} = 0.08999 \ mol \ O$

  • Divide each mole value by the smallest (0.08999) to find the mole ratio:

    • C: $\frac{0.270}{0.08999} = 3.00$

    • H: $\frac{0.3147}{0.08999} = 3.50$

    • O: $\frac{0.08999}{0.08999} = 1.00$

  • Multiply until whole number: $C<em>6H</em>7O_2$

Bond Polarity

• Question 20 involves ranking bonds in order of increasing polarity.

  • Electronegativity values: F (3.98), O (3.44), N (3.04), C (2.55).

  • The greater the difference in electronegativity between two atoms, the more polar the bond.

    • I. F-O: |3.98 - 3.44| = 0.54

    • II. F-C: |3.98 - 2.55| = 1.43

    • III. F-N: |3.98 - 3.04| = 0.94

    • IV. F-F: |3.98 - 3.98| = 0

  • Ranking in order of increasing polarity: IV < I < III < II.

Electron and Molecular Geometry

• Question 21 deals with determining the electron geometry and molecular geometry of $SF_4$.

  • $SF_4$ has 5 electron groups around the central atom (4 bonding pairs and 1 lone pair).

    • Electron geometry: Trigonal bipyramidal

    • Molecular geometry: See-saw (due to the presence of one lone pair, which distorts the shape)

Lewis Structures

• Question 22 asks to select the best Lewis structure for ClCN.

  • ClCN has a total of 16 valence electrons. The best Lewis structure should minimize formal charges.

  • A. Cl-C≡N: Cl has 3 lone pairs (6 electrons), C has no lone pairs, N has one lone pair (2 electrons).

    • Formal charges: Cl (+1), C (0), N (-1)

• Question 23, consider the partial Lewis structure (lone pairs not shown). Determine which orbitals are utilized to form the C-N bond C(sp3)-N(sp2).

• Question 28 requires determining the number of lone pairs of electrons in a Lewis dot structure.

  • There are 4 lone pairs in its Lewis dot structure.

Hybridization

• Question 29 covers the hybridizations of atoms in a Lewis structure.

  • O: sp3, C1: sp3, C2: sp, N: sp.

  • Oxygen has 4 electron groups (2 lone pairs and 2 single bonds), so it is $sp^3$ hybridized.

  • Carbon 1 has 4 single bonds, so it is $sp^3$ hybridized.

  • Carbon 2 has one single bond and one triple bond, so it is $sp$ hybridized.

  • Nitrogen has one triple bond and one lone pair, so it is $sp$ hybridized.

• Question 24 Consider the reaction of $BF<em>3$ with $NH</em>3$. First statement, the hybridization of B changes true because goes from $sp^2$ to $sp^3$. Second statement, the F-B-F bond angle changes true due the change in hybridization. Last statement, the formal charge of N changes this is also true.

Molecular Polarity

• Question 25 asks which of the following molecules are polar.

  • SF2 is polar because it has a bent shape (due to two lone pairs on S), resulting in a net dipole moment.

  • SF4 is polar because it has a see-saw shape (one lone pair on S), resulting in a net dipole moment.

  • SF6 is nonpolar because it has an octahedral shape with symmetrically arranged fluorine atoms, canceling out the dipole moments.

• Question 26 Determine the shapes of the molecules/ions has \SeO2, ClO3^-, ClO2^-. \SeO2 is bent, \ClO3^- is trigonal pyramidal, and \ClO2^- is bent.

Sigma and Pi Bonds

*Question 27 How many σ (sigma) and π (pi) bonds are present in the molecule $ClO_4^−?$

• 4 sigma bonds, 3 pi bonds are present in the molecule

Resonance Structures

• Question 30 asks about the most stable resonance structure of the perchlorate ion, ClO4–, where the formal charge of Cl is zero.

  • It has one single bond, three double bonds, and nine lone pairs.

Lattice Energy

• Question 31 asks which compound has the highest magnitude of lattice energy.

  • Lattice energy increases with increasing charge and decreasing size of ions.

    • BaCl2 vs. CaCl2: Ca2+ is smaller than Ba2+, so CaCl2 should have higher lattice energy.

    • CaCl2 vs. CaBr2: Cl– is smaller than Br–, so CaCl2 should have higher lattice energy.

    • Magnesium is more charged than Calcium. MgCl2 will have a great magnitude of lattice energy due to charge and size than CaCl2.

Electrolytes

• Question 32 asks to identify $HNO_3(aq)$.

  • Nitric acid ($HNO_3$) is a strong acid and a strong electrolyte because it completely dissociates into ions

Molecular Orbital Theory

• Question 33 asks to use the molecular orbital diagram to determine the bond order of $O_2^+$.

  • $O<em>2^+$ has 11 valence electrons. The molecular orbital configuration is $\sigma</em>{2s}^2 \sigma<em>{2s}^{2} \sigma{2p}^2 \pi{2p}^4 \pi{2p}^{</em>1}$

  • Bond order is calculated as: $\frac{bonding \ electrons - antibonding \ electrons}{2}$

  • Bond order = $\frac{8 - 3}{2} = \frac{5}{2} = 2.5$

Limiting Reactant and Mass of Product

• Question 34 asks to determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g $N<em>2O</em>4$ and 45.0 g $N<em>2H</em>4$.

  • The balanced equation is: $N<em>2O</em>4(l) + 2N<em>2H</em>4(l) \rightarrow 3N<em>2(g) + 4H</em>2O(g)$

  • Convert grams to moles:

    • $moles \ of \ N<em>2O</em>4 = \frac{50.0 g}{92.02 g/mol} = 0.543 \ mol$

    • $moles \ of \ N<em>2H</em>4 = \frac{45.0 g}{32.05 g/mol} = 1.404 \ mol$

  • Determine the limiting reactant by comparing the mole ratio to the stoichiometric ratio:

    • For every 1 mole of $N<em>2O</em>4$, 2 moles of $N<em>2H</em>4$ are required. $0.543 \ mol \ N<em>2O</em>4 \times 2 = 1.086 \ mol \ N<em>2H</em>4$. Since we have 1.404 mol $N<em>2H</em>4$, $N<em>2O</em>4$ is the limiting reactant.

  • Calculate the moles of $N_2$ formed from the limiting reactant:

    • $0.543 \ mol \ N<em>2O</em>4 \times \frac{3 \ mol \ N<em>2}{1 \ mol \ N</em>2O<em>4} = 1.629 \ mol \ N</em>2$

  • Calculate the mass of $N_2$ formed:

    • $1.629 \ mol \ N<em>2 \times \frac{28.02 g}{1 \ mol \ N</em>2} = 45.64 g \approx 45.7 \ g$

Oxidation Numbers

• Question 35 asks for the oxidation number of Cl in the ion $HClO_2$.

  • Oxidation number of H is +1.

  • Oxidation number of O is -2.

  • Let x be the oxidation number of Cl. The sum of the oxidation numbers in the ion must equal the charge of the ion.

  • $+1 + x + 2(-2) = 0$

  • $x = +3$

Percent Yield

• Question 36 The reaction of 2.0 g of sulfur with 2.0 g of oxygen according to the reaction below produced 1.60 g of sulfur trioxide. The percent yield of $SO<em>3$ was %.

  • $2S + 3O<em>2 \rightarrow 2SO</em>3$

  • First determine the limiting reactant by dividing the number of moles of each substance by its coefficient.

    • $S \times \frac{1 mol}{32.06 g} = 0.0624$; dividing by coefficient of 2 yields .0312

    • $O_2 \times \frac{1 mol}{32 g} = 0.0625$; dividing by coefficient of 3 yields .0208. Thus oxygen is the LR.

  • Now determine the theoretical yield.

    • $0.0625 mol \times 80.06 g = 5.004 g$

  • \% yield is calculating by dividing $Actual/Theoretical \times100$.

    • $1.6/5.004 \times 100 = 32 \%$

Molarity

• Question 37 asks to calculate the concentration of $FeCl_3$ in a solution.

  • Molarity(M) = $\frac{moles \ of \ solute}{Liters \ of \ solution}$

  • $MW \ FeCl_3 = 162.2 \frac{g}{mol}$

  • $moles \ FeCl_3 = \frac{30 \text{grams}}{162.2 \frac{g}{mol}} = .185$ moles

  • $M = \frac{.185}{.275} = .673 M$

Specific Heat

• Question 38 calculating for heat lost and gained \Delta Q = mc\DeltaT for metal and water, where c is specific heat.

  • Heat lost by metal = Heat gained by water

  • (mc\DeltaT){metal} = (mc\DeltaT){water}

  • $(18.5)(c)(77.5-26.1) = (58.5)(4.18)(26.1 - 24.0)$

  • c = .54$\frac{J}{g^oC}$

Neutralization Reaction

• Question 39 Neutralization of 27.28 mL $H<em>2SO</em>4$ required 10.87 mL of 0.28 M NaOH.

• $H<em>2SO</em>4(aq) + 2NaOH(aq) \rightarrow Na<em>2SO</em>4(aq) + 2H_2O(l)$

• Determine the molarity of $H<em>2SO</em>4$.

• At equivalence, moles of $H<em>2SO</em>4$ times 2 (because it is diprotic) = moles of NaOH.

• $M<em>A V</em>A = M<em>B V</em>B$

• Molarity of acid = $1. 005M$

Net Ionic Equation

• Question 40 asks for the net ionic equation for the reaction (if any) that occurs when aqueous solutions of $K<em>2S$ and $Fe(NO</em>3)_2$ are mixed.

  • When $K<em>2S(aq)$ and $Fe(NO</em>3)<em>2(aq)$ are mixed, the possible products are $KNO</em>3$ and $FeS$.

  • According to solubility rules, all nitrate compounds and all compounds containing alkali metal ions are soluble. $FeS$ is insoluble.

  • The net ionic equation is: $Fe^{2+}(aq) + S^{2-}(aq) \rightarrow FeS(s)$

Reducing Agent

• Question 41 Select the reducing agent from the equation, $IO<em>4 − + 7 I− + 8 H+ \rightarrow 4 I</em>2 + 4 H_2O$.

  • The reducing agent is the substance that is oxidized (loses electrons). In this case, $I^−$ is oxidized to $I_2$.

Molar Mass Calculation

*Question 42 A 1.40 g sample of H2X reacts completely with 30.0 mL of 0.400 M NaOH according to reaction: H2X(aq) + 2NaOH(aq) → 2H2O(l) + 2NaX(aq). What is the molar mass of H2X?
*Moles of $NaOH = M \times V = 0.400 \times0.030 = .012$ moles of NaOH.
*From the balanced equation, the mole ratio of $H<em>2X$ to NaOH is 1:2. Moles of $H</em>2X = 0.012/2 = 0.006$ moles.
*Molar mass of $$H_2